Symmetrical component transformation

Three Phase System

Balanced and Unbalanced faults

Symmetrical Components

Vector Presentation

Matrix Relations

Power plant Transformer Switching station

Line Load

Three-phase system

Single-line diagram

Equivalent single-phase circuit

•Balanced Cases

– three-phase fault

– (symmetrical) load flow

•Unbalanced Cases

– Single line to ground fault

– Line to line fault

– Double line to ground fault

– (unsymmetrical load flow)

◦Unbalanced faults in power systems require a phase by phase solution method or other techniques.

◦One of the most useful techniques to deal with unbalanced networks is the “symmetrical component” method, developed in 1918 by C.L. Fortescue.

•Reasons for use of symmetrical component –Unbalanced systems are difficult to handle

• -> several independent balanced systems are

easier to handle than one unbalanced system

– Transformation of one unbalanced 3-phase system into 3 balanced 3-phase systems.

• -> for each system only one phase has to be

considered

Any three unbalanced set of voltages or currents can be resolved into three balanced systems of voltages or currents, referred to as the system symmetrical components, defined as follows:

Positive Sequence components: three phasors of equal magnitude displaced 120 degrees from each other following the positive sequence

Negative Sequence components: three phasors of equal magnitude displaced 120 degrees of each other following the negative sequence

Zero Sequence components: three parallel phasors having equal magnitude and angle

For a 3-ph system: 3 unbalanced phasors can be resolved into 3 balanced systems of 3 phasors each

Let Va, Vb, Vc be the Phase voltages According to Fortescue, these can be transformed into Positive-seq. voltages: Va1, Vb1, Vc1 Negative-seq. voltages: Va2, Vb2, Vc2 zero-sequence voltages: Va0, Vb0, Vc0 Thus, Va = Va1 + Va2 + Va0 Vb = Vb1 + Vb2 + Vb0 Vc = Vc1 + Vc2 + Vc0

Va1

Vb1

Vc1

Va2

Vc2

Vb2

Va1

Va2 Va0

Va Vc

Vb

Va0 Vb0 Vc0

The ‘a’ operator a = 1<1200 = -0.5 + j 0.866

a I rotates I by 1200

a2 = 1<2400 = -0.5 – j 0.866

a3 = 1<3600 = 1<00 = 1 + j 0

1 + a + a2 = 0 1

a2

a

-a

-a2

-1

From figure previous figures

Vb1 = a2Va1 Vc1 = a Va1

Vb2 = a Va2 Vc2 = a2 Va2

Vb0 = Va0 Vc0 = Va0

sub. In Eq. (Slide 8) we get:

Thus, Va = Va0 + Va1 + Va2

Vb = Va0 + a2Va1 + a Va2

Vc = Va0 + a Va1 + a2Va2

Matrix Relations

Let

And Inverse of A is

a a0

2

b a1

2

c a2

v v 1 1 1

Vp = v ; Vs = v ; A = 1 a a

v v 1 a a

-1 213

2

1 1 1

A = 1 a a

1 a a

Matrix Relations

Similarly currents can be obtained using their

symmetrical components

2

1

0

2

2

1

1

111

a

a

a

c

b

a

v

v

v

aa

aa

v

v

v

Matrix Relations

Vp = A Vs; Vs = A-1Vp

Va0 = 1/3 (Va + Vb + Vc)

Va1 = 1/3 (Va + aVb + a2Vc)

Va2 = 1/3 (Va + a2Vb + aVc)

Matrix Relations

c

b

a

a

a

a

v

v

v

aa

aa

v

v

v

2

2

31

2

1

0

1

1

111